Hey there!
<h3>"Expanded form" is basically, setting it up in a addition form to give you that result</h3>
![4 = 400,000 \\ 0 = 0 \\ 3 = 3,000 \\ 8 = 800 \\ 9 = 90 \\ 2 =2 \\ \\ \\ \\ Answer: 400,000 + 0 + 3,000 + 800 + 90 + 2](https://tex.z-dn.net/?f=%204%20%3D%20400%2C000%20%5C%5C%20%200%20%3D%200%20%5C%5C%203%20%3D%203%2C000%20%20%5C%5C%208%20%3D%20800%20%5C%5C%209%20%3D%2090%20%5C%5C%202%20%3D2%20%5C%5C%20%5C%5C%20%5C%5C%20%5C%5C%20Answer%3A%20%20400%2C000%20%2B%200%20%2B%203%2C000%20%2B%20800%20%2B%2090%20%2B%202%20%20)
Good luck on your assignment and enjoy your day!
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12x - 4x = 8 + 3 - 3
8x = 8
X = 1
4^-3
You first need to make the exponent positive. To do this put it in a fraction.
1/4^3
Now solve the exponent.
4 * 4 * 4 =
^
16*4 = 64
1/64 is your answer
19g/1000 would be the right answer when simplified.
The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
![\sum{a+b}=\sum{a}+\sum{b}](https://tex.z-dn.net/?f=%5Csum%7Ba%2Bb%7D%3D%5Csum%7Ba%7D%2B%5Csum%7Bb%7D)
![\sum{kx}=k\sum{x}](https://tex.z-dn.net/?f=%5Csum%7Bkx%7D%3Dk%5Csum%7Bx%7D)
![\sum_{r=1}^n{1}=n](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7B1%7D%3Dn)
![\sum_{r=1}^n{r}=n(n+1)/2](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7Br%7D%3Dn%28n%2B1%29%2F2)
Armed with the above rules, we can split up the summation into simple terms:
![\sum_{r=1}^n{40r-21n+8}=n](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7B40r-21n%2B8%7D%3Dn)
![=40\sum_{r=1}^n{r}-21n\sum_{r=1}^n{1}+8\sum_{r=1}^n{1}](https://tex.z-dn.net/?f=%3D40%5Csum_%7Br%3D1%7D%5En%7Br%7D-21n%5Csum_%7Br%3D1%7D%5En%7B1%7D%2B8%5Csum_%7Br%3D1%7D%5En%7B1%7D)
![=40\frac{n(n+1)}{2}-21n^2+8n](https://tex.z-dn.net/?f=%3D40%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D-21n%5E2%2B8n)
![=20n(n+1)-21n^2+8n](https://tex.z-dn.net/?f=%3D20n%28n%2B1%29-21n%5E2%2B8n)
![=28n-n^2](https://tex.z-dn.net/?f=%3D28n-n%5E2)
=> (a)
f(x)=28n-n^2=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b)
f(x) is a maximum when n=14
(c)
the maximum value of f(x) is f(14)=196